Bunuel wrote:

In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1

B. \(\frac{\pi}{2}\)

C. \(\frac{\pi}{3}\)

D. \(\frac{3\pi}{8}\)

E. \(\frac{4\pi}{9}\)

Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:This problem is based on a famous theorem of

Hippocrates of Chios (c. 470 – c. 410 BCE), a predecessor of Euclid. The easiest way to see this is as follows.

Let AC = BC = 2S. This is the radius of the larger circle, which has an area of 4pi*S^2, so that the quarter circle:

must have an area of pi*S^2. Hold that thought.

Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be

\(AB=2\sqrt{2}S\)

and the radius of the smaller circle is

\(AM=\sqrt{2}S\)

The area of that whole circle would be 2pi*S^2, and the area of the semicircle:

would be half of that, pi*S^2. Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area. That’s a very big idea!

Now, look at the circular segment:

We don’t need to know that area: simply call it J.

If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle:

pi*S^2 – J = area of triangle ABC

If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune:

pi*S^2 – J = area of the lune

Because those two areas equal the same thing, they must equal each other.

area of triangle ABC = area of the lune

The two are equal, so their ratio is 1.

Answer = (A)

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